A motorist traveling along a straight highway at a constant speed of 30\text{ m/s} observe… — JAMB UTME

A motorist traveling along a straight highway at a constant speed of $30\text{ m/s}$ observes a road hazard ahead. The driver experiences a reaction delay of $0.5\text{ s}$ before applying the brakes, after which the vehicle decelerates uniformly at a rate of $5\text{ m/s}^2$ . What is the total distance traveled by the vehicle from the instant the hazard is observed until the vehicle comes to a complete stop?

A
$90\text{ m}$
B
$97.5\text{ m}$
$105\text{ m}$ Answer
D
$195\text{ m}$

Answer

The total distance traveled by the vehicle before coming to a stop is

105\text{ m}

The motion consists of two distinct stages: a constant-speed phase during the

0.5\text{ s}

reaction time (

15\text{ m}

) and a uniformly decelerating phase until rest (

90\text{ m}

). Summing both components yields

105\text{ m}

Step-by-Step Solution

Calculate the reaction distance (

s_1

) covered during the driver's reaction delay.

s_1 = u \times t_r = 30\text{ m/s} \times 0.5\text{ s} = 15\text{ m}

Before the brakes are applied, the vehicle continues moving at its initial constant speed of

30\text{ m/s}

Calculate the braking distance (

s_2

) using the third equation of motion.

v^2 = u^2 + 2as_2 \implies 0^2 = 30^2 + 2(-5)s_2 \implies 10s_2 = 900 \implies s_2 = 90\text{ m}

The vehicle decelerates from

30\text{ m/s}

to a final velocity of

0\text{ m/s}

at a uniform deceleration rate of

a = -5\text{ m/s}^2

Sum the reaction distance and the braking distance to find the total stopping distance.

s_{\text{total}} = s_1 + s_2 = 15\text{ m} + 90\text{ m} = 105\text{ m}

The total stopping distance is the sum of the distance covered before braking begins and the distance covered while braking.

Key Concept

Two-phase linear motion combining constant velocity reaction distance and uniform deceleration braking distance.

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